Unizor – Creative Mind through Art of Mathematics: Trigonometry+ 09: UNIZOR.COM – Math+ & Problems

Notes to a video lecture on http://www.unizor.com Trigonometry+ 09Problem AProve:limx→0[sin(x)/x] = 1Proof AIn the lecture Trigonometry 07 (Problem C) of this Math+ & Problems course we have proven geometrically that for an acute angle θ, measured in radians, the following inequalities are true:sin(θ) ≤ θ ≤ tan(θ)The left inequality can be transformed intosin(x)/x ≤ 1Using the definition of function tan(x)=sin(x)/cos(x), the right inequality can be transformed intocos(x) ≤ sin(x)/xCombining together, these inequalities givecos(x) ≤ sin/x ≤ 1As x→0, lower and upper boundaries of sin(x)/x tend to 1.Therefore, limx→0[sin(x)/x] = 1A different proof can be obtained based on using the L’Hopital’s Rule (see UNIZOR.COM: Math 4 Teens – Calculus – Derivatives – Main Theorems – L’Hopital’s Rule) as follows:limx→0[sin(x)/x] == limx→0[sin'(x)/x’] == limx→0[cos(x)/1] = 1The following theorems assume that the overall dimensions of regular polygons are not growing to infinity but restricted to some values, while the number of sides of polygons increases to infinity.Problem BProve that the perimeter of a regular polygon tends to a circumference of an inscribed into it circle, if the number of its sides tends to infinity.More precisely, prove that the difference between a perimeter of a regular N-sided polygon and a circumference of a circle inscribed into it is an infinitesimal variable as N→∞.Proof BLetp – perimeter of a regular polygon,N – number of its sides,r – radius of an inscribed circle.Then,p = 2N·r·tan(½·2π/N) == 2N·r·sin(π/N)/cos(π/N) =(substitute x=π/N,so if N→∞, x→0)= 2π·r·[sin(x)/x]/cos(x)As x→0, cos(x)→1.As proven in Problem A above, limx→0[sin(x)/x] = 1Therefore, the limit of a perimeter p equals to 2π·r, which is a circumference of an inscribed circle.Problem CProve that the perimeter of a regular polygon tends to a circumference of a circumscribed around it circle, if the number of its sides tends to infinity.More precisely, prove that the difference between a perimeter of a regular N-sided polygon and a circumference of a circle circumscribed around it is an infinitesimal variable as N→∞.Proof CLetp – perimeter of a regular polygon,N – number of its sides,R – radius of a circumscribed circle.Then,p = 2N·R·sin(½·2π/N) == 2N·R·sin(π/N) =(substitute x=π/N,so if N→∞, x→0)= 2π·R·[sin(x)/x]As x→0, sin(x)/x→1.Therefore, the limit of a perimeter p equals to 2π·R – a circumference of a circumscribed circle.Problem DProve that the area of a regular polygon tends to an area of an inscribed into it circle, if the number of its sides tends to infinity.More precisely, prove that the difference between an area of a regular N-sided polygon and an area of a circle inscribed into it is an infinitesimal variable as N→∞.Proof DLetA – area of a regular polygon,p – perimeter of a regular polygon,N – number of its sides,r – radius of an inscribed circle.Dividing a regular polygon into N equilateral triangles with common vertex in a center of an inscribed circle leads to an obvious formula for its areaA = ½·p·rSince the limit of a perimeter p equals to 2π·r,limN→∞A = ½·2π·r·r = π·r²Problem EProve that the area of a regular polygon tends to an area of a circumscribed around it circle, if the number of its sides tends to infinity.More precisely, prove that the difference between an area of a regular N-sided polygon and an area of a circle circumscribed around it is an infinitesimal variable as N→∞.Proof ELetA – area of a regular polygon,p – perimeter of a regular polygon,N – number of its sides,R – radius of a circumscribed circle.r – radius of an inscribed circle.Notice thatr² + (p/2N)² = R²and, therefore,limN→∞(R−r) = 0.That is, radiuses of inscribed and circumscribed circles have the same limit.Dividing a regular polygon into N equilateral triangles with common vertex in a center of a circumscribed circle leads to an obvious formula for its areaA = ½·p·√R²−(p/2N)²In this formula, as proved above, the limit of a perimeter p equals to 2π·r, which is the same as 2π·R.An expression p/2N has limit zero as N→∞.Therefore,limN→∞A = ½·2π·R·R = π·R²